Chapter 2.6, Problem 7E

To determine

To find: The equation of the tangent line to the curve at the given point.

Answer to Problem 7E

The equation of the tangent line to the curve $y=\sqrt{x}$ at the point (1, 1) is $y=\frac{1}{2}x+\frac{1}{2}$.

Explanation of Solution

Given:

The equation of the curve is $y=\sqrt{x}$.

The curve passing through the point (1, 1).

Formula used:

The slope of the tangent curve $y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$ is,

$m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ (1)

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is,

$y-f\left(a\right)=f^{\prime }\left(a\right)\left(x-a\right)$ (2)

Difference of square formula: $\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$

Calculation:

Obtain the slope of the tangent line to the parabola at the point $\left(1,1\right)$.

Substitute $a=1$ and $f\left(a\right)=1$ in equation (1),

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{\left(\sqrt{x}\right)-1}{x-1}\\ =\underset{x\to 1}{\lim }\frac{\sqrt{x}-1}{x-1}\end{array}$

Multiply both the numerator and denominator by the conjugate of the numerator.

$m=\underset{x\to 1}{\lim }\left(\frac{\sqrt{x}-1}{x-1}\times \frac{\sqrt{x}+1}{\sqrt{x}+1}\right)$

Apply the difference of squares formula,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{\left({\left(\sqrt{x}\right)}^2-{\left(1\right)}^2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\\ =\underset{x\to 1}{\lim }\frac{\left(x-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\end{array}$

Since the limit x approaches 1 but is not equal to 1, cancel the common term $\left(x-1\right)\left(\ne 0\right)$ from both the numerator and the denominator,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{1}{\left(\sqrt{x}+1\right)}\\ =\frac{1}{\left(\sqrt{1}+1\right)}\\ =\frac{1}{1+1}\\ =\frac{1}{2}\end{array}$

Thus, the slope of the tangent line to the curve at the point (1, 1) is $\underset{\_}{m=\frac{1}{2}}$.

Obtain the equation of the tangent line.

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of f at a, which is $f^{\prime }\left(a\right)=\frac{1}{2}$.

Substitute $a=1,$$f\left(a\right)=1$ and $f^{\prime }\left(a\right)=\frac{1}{2}$ in equation (2),

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-\left(1\right)=\frac{1}{2}\left(x-1\right)\\ y-1=\frac{1}{2}x-\frac{1}{2}\end{array}$

Isolate y as shown below.

$\begin{array}{c}y=\frac{1}{2}x-\frac{1}{2}+1\\ =\frac{1}{2}x+\frac{1}{2}\end{array}$

Thus, the equation of the tangent line is $y=\frac{1}{2}x+\frac{1}{2}$.